3 Marks Question

Question 13 Marks

The density of water at 0°C is $0.998g\ cm^{-3}$ and at 4^\circC is $1000g\ cm^{-3}$.Calculate the average coefficient of volume expansion of water in the temperature range 0 to 4°C.

Answer

$\text{f}_{0^\circ\text{C}}=0.098\text{g/m}^3$$\text{f}_{4^\circ\text{C}}=1\text{g/m}^3$
$\text{f}_{0^\circ\text{C}}=\frac{\text{f}_{4^\circ\text{C}}}{1+\gamma\Delta\text{T}}$
$\Rightarrow0.998=\frac{1}{1+\gamma\times4}$
$\Rightarrow1+4\gamma=\frac{1}{0.998}$
$\Rightarrow4+\gamma=\frac{1}{0.998}-1$
$\Rightarrow\gamma=0.0005\approx5\times10^{-4}$
As density decreases $\gamma=-5\times10^{-4}$

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Question 23 Marks

If mercury and glass had equal coefficient of volume expansion, could we make a mercury thermometer in a glass tube?

Answer

Yes, we can make a mercury thermometer in a glass tube. Mercury and glass have equal coefficients of volume expansion. So, when temperature changes, the increase in the volume of the glass tube as which is equal to the real increase in volume minus the increase in the volume of the container, would be zero. Hence, it will give correct reading at every temperature.

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Question 33 Marks

A concrete slab has a length of 10m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is $35^\circ C.$ The coefficient of linear expansion of concrete is ${1.0 \times 10^{-5}}^\circ C^{-1}$.

Answer

$\text{L}_1=?,$
$\text{L}_0=10\text{m},$
$\alpha=1\times10^{-5}/^\circ\text{C},$
$\text{t}=35$
$\text{L}_1=\text{L}_0(1+\alpha\text{t})$
$=10\Big(1+10^{-5}\times35\Big)$
$=10+35\times10^{-4}$
$=10.0035\text{m}$

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Question 43 Marks

Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature. Coefficients of linear expansion of iron and aluminium are ${12 \times 10^{-6}} ^\circ C^{-1}$ and ${23 \times 10^{-6}}^\circ C^{-1}$ respectively.

Answer

Iron rod $(L_{Fe})$ $\alpha_{\text{Fe}}=12\times18^{-8}/^\circ\text{C}$
Aluminium rod $(L_{Al})$
$\alpha_\text{Al}=23\times10^{-8}/^\circ\text{C}$
Since the difference in length is independent of temp. Hence the different always remains constant.
$\text{L'}_{\text{Fe}}=\text{L}_{\text{Fe}}(1+\alpha_{\text{Fe}}\times\Delta\text{T}) \ ...(1)$
$\text{L'}_\text{Al}=\text{L}_\text{Al}(1+\alpha_{\text{Al}}\times\Delta\text{T})\ ...(2)$
$\text{L'}_\text{Fe}-\text{L'}_\text{Al}$
$=\text{L}_\text{Fe}-\text{L}_\text{Al}+\text{L}_\text{Fe}\times\alpha_\text{Fe}\times\Delta\text{T}-\text{L}_\text{Al}\times\alpha_\text{Al}\times\Delta\text{T}$
$\frac{\text{L}_\text{Fe}}{\text{L}_\text{Al}}=\frac{\alpha_\text{Al}}{\alpha_\text{fe}}$
$=\frac{23}{12}=23:12$

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Question 53 Marks

A circular hole of diameter 2.00cm is made in an aluminium plate at $0^\circ C$. What will be the diameter at $100^\circ C?$ $\alpha$ for aluminium$= {2.3 \times 10^{-5}}{^\circ C^{-1}}.$

Answer

$d_1 = 2cm = 2 \times 10^{-2}$
$t_1 = 0^\circ C,$
$t_2 = 100^\circ C$
$\alpha_\text{al}=2.3\times10^{-5}/^\circ\text{C}$
$\text{d}_2=\text{d}_1(1+\alpha\Delta\text{t})=2\times10^{-2}(1+2.3\times10^{-5}10^2)$
$=0.02+0.000046=0.020046\text{m}=2.0046\text{cm}$

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Question 63 Marks

A metre scale made of steel is calibrated at 20°C to give correct reading. Find the distance between the 50cm mark and the 51cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is $1.1 \times 10^{-5} $$ ^\circ C^{-1}.$

Answer

$t_1 = 20$$^\circ C, t_2 = 10$$^\circ C, L_1 = 1cm = 0.01m, L_2 =? \alpha_{\text{steel}}=1.1\times10^{-5}/^\circ\text{C}$
$\text{L}_2=\text{L}_1(1+\alpha_\text{steel}\Delta\text{T})$
$=0.01(1+101\times10^{-5}\times10)$
$=0.01+0.01\times1.1\times10^{-4}$
$=10^4\times10^{-6}+1.1\times10^{-6}$
$=10^{-6}(10000+1.1)=10001.1$ $=1.00011\times10^{-2}\text{m}=1.00011\text{cm}$

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Question 73 Marks

The length of a brass rod is found to be smaller less on a hot summer day than on a cold winter day as measured by the same aluminium scale. Do we conclude that brass shrinks on heating?

Answer

On a hot summer day, metals tend to expand due to the heat. Different metals have different expansion coefficients. The coefficient of linear expansion of aluminium is more than that of brass. Therefore, it'll expand more than brass, leading to an apparent decrease in length of the brass rod, as measured by the aluminium scale. So, we cannot conclude that brass shrinks on heating. Instead, aluminium expands more than brass on heating.

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Question 83 Marks

In defining the ideal gas temperature scale, it is assumed that the pressure of the gas at constant volume is proportional to the temperature T. How can we verify whether this is true or not? Are we using the kinetic theory of gases? Are we using the experimental result that the pressure is proportional to temperature?

Answer

The ideal gas thermometer is based on the ideal gas equation, PV = nRT, where P is pressure of the gas at constant volume V with n number of moles at temperature T.
Therefore, P = constant × T. According to this relation, if the volume of the gas used is constant, the pressure will be directly proportional to the temperature of the gas. We need not use kinetic theory of gases or any experimental results.

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Question 93 Marks

Does the temperature of a body depend on the frame from which it is observed?

Answer

No, the temperature of a body is not dependent on the frame from which it is observed. This is because atoms/ molecules of matter move or vibrate in all possible directions. Increase in velocity at a particular direction of the container/ matter does not increase or decrease the overall velocity of the molecules/ atoms because of the random collisions the entities suffer. So, there is no net rise in temperature of the system.

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Question 103 Marks

A railway track (made of iron) is laid in winter when the average temperature is 18°C. The track consists of sections of 12.0m placed one after the other. How much pp should be left between two such sections 80 that there is no compression during summer when the maximum temperature goes to 48°C? Coefficient of linear expansion of iron ${= 11 \times 10^{-6}}^\circ C^{-1}.$

Answer

$\text{L}_0=12\text{cm},$
$\alpha=11\times10^{-6}/^\circ\text{C}$
$\text{tw}=18^\circ\text{C}$
$\text{ts}=48^\circ\text{C}$
$\text{Lw}=\text{L}_0(1+\alpha\text{tw})=12(+11\times10^{-5}\times18)=12.002376\text{m}$
$\text{Ls}=\text{L}_0(1+\alpha\text{ts})=12(1+11\times10^{-5}\times48)=12.006336\text{m}$
$\Delta\text{L}=12.006336-12.002376=0.00396\text{m}\approx0.4\text{cm}$

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Question 113 Marks

Is it possible for two bodies to be in thermal equilibrium if they are not in contact?

Answer

Two bodies are said to be in thermal equilibrium if they are at the same temperature. Consider two bodies A and B that are not in contact with each other but in contact with a heat reservoir. Since both the bodies will attain the temperature of the reservoir, they will be at the same temperature and, hence, in thermal equilibrium. Therefore, it is possible to have two bodies in thermal equilibrium even though they are not in contact.

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Question 123 Marks

A steel rod is rigidly clamped at its two ends. The rod is under zero tension at 20°C. If the temperature rises to 100°C, what force will the rod exert on one of the clampa? Area of Cross section of the rod $= 2.00mm^2.$ Coefficient of linear expansion of steel ${12.0 \times 10^{-6} }^\circ C^{-1}$ and Young's modulus of steel $= 2.00 \times 10^{11}N\ m^{-2}.$

Answer

$\theta_1=20^\circ\text{C}$
$\theta_2=100^\circ\text{C}$
$\text{A}=2\text{mm}^2=2\times10^{-6}\text{m}^2$
$\alpha_\text{steel}=12\times10^{-6}\ /^\circ\text{C},$
$\text{Y}_\text{steel}=12\times10^{11}\text{N/m}^2$
Force exerted on the clamps = ?
$\frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Strain}}=\text{Y}$
$\Rightarrow\text{F}=\frac{\text{Y}\times\Delta\text{L}}{\text{L}}\times\text{L}$
$=\frac{\text{YL}\alpha\Delta\theta\text{A}}{\text{L}}=\text{YA}\alpha\Delta\theta$
$=2\times10^{11}\times2\times10^{-6}\times12\times10^{-6}\times80=384\text{N}$

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